Ir al contenido principal

## Due April 5

### Problem 1

1. The set $L^1(\mathbb R^d)$ of integrable functions is a complex vector space, and $f\mapsto \int f$ is a linear functional.
2. $\displaystyle ||f||_1 = \int f$ is a norm on $L^1(\mathbb R^d)$, when $L^1(\mathbb R^d)$ is seen as a set of equivalence classes of $f\sim g$ if and only if $f=g$ a.e.

### Problem 2

Let $L^2(\mathbb R^d)$ be the set of measurable functions $f$ such that $\displaystyle \int |f|^2 < \infty$, seen as a set of equivalence classes of $f\sim g$ if and only if $f=g$ a.e.
1. $L^2(\mathbb R^d)$ is a complex vector space.
2. The bilinear form $\displaystyle \langle f, g \rangle = \int f \bar g$ is well defined on $L^2(\mathbb R^d)$ and is an inner product.
3. $L^2(\mathbb R^d)$ is complete with the norm $\displaystyle ||f||_2 = \sqrt{\langle f, f \rangle}.$ (Hint: Proceed as in the case of $L^1$ seen in class.)

### Problem 3

1. If $f\in L_\text{loc}^1(\mathbb R^d)$, then $Mf$ might be infinite at every point.
2. If $f\in L^1(\mathbb R^d)$, then $Mf$ might be infinite at some points.

### Problem 4

We say that $\{K_t\}_{t>0}$ is a family of better kernels if it satisfies
• $\displaystyle \int_{\mathbb R^d} K_t(x) dx = 1$ for all $t>0$;
• there exists $A>0$ such that $|K_t(x)| \le \dfrac{A}{t^d}$ for all $x\in\mathbb R^d$ and $t>0$; and
• there exists $A'>0$ such that $|K_t(x)| \le \dfrac{A't}{|x|^{d+1}}$ for all $x\in\mathbb R^d$ and $t>0$.
1. If $\Phi\in L^1(\mathbb R^d)$ and $|\Phi(x)|\le A/(1 + |x|)^{d+1}$, then its dilations $\{\Phi_t\}_{t>0}$ form a family of better kernels.
2. If $\{K_t\}_{t>0}$ is a family of better kernels, then there exists a constant $c>0$ such that, if $f\in L^1(\mathbb R^d)$, $|K_t*f(x)| \le cMf(x)$ for all $x\in\mathbb R^d$ and $t>0$.
3. If $\{K_t\}_{t>0}$ is a family of better kernels and $f\in L^1(\mathbb R^d)$, then $\displaystyle \lim_{t\to0} K_t*f(x) = f(x)$ for almost every $x\in\mathbb R^d$.

### Problem 5

Let $\{K_t\}_{t>0}$ be a family of better kernels and $\theta>0$.
1. There exists a constant $c_\theta>0$ such that, for $f\in L^1(\mathbb R^d)$ and $x\in\mathbb R^d$, $|K_t*f(y)| \le c_\theta Mf(x)$ for all $(y,t)\in \Gamma_\theta(x)$.
2. If $f\in L^1(\mathbb R^d)$, then $\displaystyle \lim_{\substack{(y,t)\to(x,0)\\(y,t)\in \Gamma_\theta(x)}} K_t*f(y) = f(x)$ for almost every $x\in \mathbb R^d$.